S spezialize New member Joined Sep 27, 2005 Messages 26 Nov 3, 2005 #1 I need to find the linearization for L(x) for f(x)= sqrt(x+1) at x=3.. can anyone give me some steps on how to go about for solving this.
I need to find the linearization for L(x) for f(x)= sqrt(x+1) at x=3.. can anyone give me some steps on how to go about for solving this.
pka Elite Member Joined Jan 29, 2005 Messages 11,978 Nov 3, 2005 #2 Well \(\displaystyle L(x) = f(a) + f'(a)(x - a)\) here a=3. Also \(\displaystyle f(x) = \sqrt {x + 1} ,\quad f'(x) = \frac{1}{{2\sqrt {x + 1} }}\)
Well \(\displaystyle L(x) = f(a) + f'(a)(x - a)\) here a=3. Also \(\displaystyle f(x) = \sqrt {x + 1} ,\quad f'(x) = \frac{1}{{2\sqrt {x + 1} }}\)
S spezialize New member Joined Sep 27, 2005 Messages 26 Nov 3, 2005 #3 so i just plug in 3 into that?
pka Elite Member Joined Jan 29, 2005 Messages 11,978 Nov 3, 2005 #4 \(\displaystyle f(3) = \sqrt {3 + 1} = 2\) \(\displaystyle f'(3) = \frac{1}{{2\sqrt {3 + 1} }} = \frac{1}{4}\) \(\displaystyle L(x) = 2 + \frac{1}{4}(x - 3)\)
\(\displaystyle f(3) = \sqrt {3 + 1} = 2\) \(\displaystyle f'(3) = \frac{1}{{2\sqrt {3 + 1} }} = \frac{1}{4}\) \(\displaystyle L(x) = 2 + \frac{1}{4}(x - 3)\)